Base ten decimal number 0.000 38 converted to 32 bit single precision IEEE 754 binary floating point standard

How to convert the decimal number 0.000 38(10)
to
32 bit single precision IEEE 754 binary floating point
(1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. First, convert to binary (base 2) the integer part: 0. Divide the number repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:

  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

2. Construct the base 2 representation of the integer part of the number, by taking all the remainders starting from the bottom of the list constructed above:

0(10) =


0(2)

3. Convert to binary (base 2) the fractional part: 0.000 38. Multiply it repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:

  • #) multiplying = integer + fractional part;
  • 1) 0.000 38 × 2 = 0 + 0.000 76;
  • 2) 0.000 76 × 2 = 0 + 0.001 52;
  • 3) 0.001 52 × 2 = 0 + 0.003 04;
  • 4) 0.003 04 × 2 = 0 + 0.006 08;
  • 5) 0.006 08 × 2 = 0 + 0.012 16;
  • 6) 0.012 16 × 2 = 0 + 0.024 32;
  • 7) 0.024 32 × 2 = 0 + 0.048 64;
  • 8) 0.048 64 × 2 = 0 + 0.097 28;
  • 9) 0.097 28 × 2 = 0 + 0.194 56;
  • 10) 0.194 56 × 2 = 0 + 0.389 12;
  • 11) 0.389 12 × 2 = 0 + 0.778 24;
  • 12) 0.778 24 × 2 = 1 + 0.556 48;
  • 13) 0.556 48 × 2 = 1 + 0.112 96;
  • 14) 0.112 96 × 2 = 0 + 0.225 92;
  • 15) 0.225 92 × 2 = 0 + 0.451 84;
  • 16) 0.451 84 × 2 = 0 + 0.903 68;
  • 17) 0.903 68 × 2 = 1 + 0.807 36;
  • 18) 0.807 36 × 2 = 1 + 0.614 72;
  • 19) 0.614 72 × 2 = 1 + 0.229 44;
  • 20) 0.229 44 × 2 = 0 + 0.458 88;
  • 21) 0.458 88 × 2 = 0 + 0.917 76;
  • 22) 0.917 76 × 2 = 1 + 0.835 52;
  • 23) 0.835 52 × 2 = 1 + 0.671 04;
  • 24) 0.671 04 × 2 = 1 + 0.342 08;
  • 25) 0.342 08 × 2 = 0 + 0.684 16;
  • 26) 0.684 16 × 2 = 1 + 0.368 32;
  • 27) 0.368 32 × 2 = 0 + 0.736 64;
  • 28) 0.736 64 × 2 = 1 + 0.473 28;
  • 29) 0.473 28 × 2 = 0 + 0.946 56;
  • 30) 0.946 56 × 2 = 1 + 0.893 12;
  • 31) 0.893 12 × 2 = 1 + 0.786 24;
  • 32) 0.786 24 × 2 = 1 + 0.572 48;
  • 33) 0.572 48 × 2 = 1 + 0.144 96;
  • 34) 0.144 96 × 2 = 0 + 0.289 92;
  • 35) 0.289 92 × 2 = 0 + 0.579 84;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (losing precision...)

4. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the constructed list above:

0.000 38(10) =


0.0000 0000 0001 1000 1110 0111 0101 0111 100(2)

Positive number before normalization:

0.000 38(10) =


0.0000 0000 0001 1000 1110 0111 0101 0111 100(2)

5. Normalize the binary representation of the number, shifting the decimal mark 12 positions to the right so that only one non zero digit remains to the left of it:

0.000 38(10) =


0.0000 0000 0001 1000 1110 0111 0101 0111 100(2) =


0.0000 0000 0001 1000 1110 0111 0101 0111 100(2) × 20 =


1.1000 1110 0111 0101 0111 100(2) × 2-12

Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign: 0 (a positive number)


Exponent (unadjusted): -12


Mantissa (not normalized): 1.1000 1110 0111 0101 0111 100

6. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2:

Exponent (adjusted) =


Exponent (unadjusted) + 2(8-1) - 1 =


-12 + 2(8-1) - 1 =


(-12 + 127)(10) =


115(10)


  • division = quotient + remainder;
  • 115 ÷ 2 = 57 + 1;
  • 57 ÷ 2 = 28 + 1;
  • 28 ÷ 2 = 14 + 0;
  • 14 ÷ 2 = 7 + 0;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

Exponent (adjusted) =


115(10) =


0111 0011(2)

7. Normalize mantissa, remove the leading (the leftmost) bit, since it's allways 1 (and the decimal point, if the case) then adjust its length to 23 bits, only if necessary (not the case here):

Mantissa (normalized) =


1. 100 0111 0011 1010 1011 1100 =


100 0111 0011 1010 1011 1100

Conclusion:

The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)


Exponent (8 bits) =
0111 0011


Mantissa (23 bits) =
100 0111 0011 1010 1011 1100

Number 0.000 38, a decimal, converted from decimal system (base 10)
to
32 bit single precision IEEE 754 binary floating point:


0 - 0111 0011 - 100 0111 0011 1010 1011 1100

(32 bits IEEE 754)
  • Sign (1 bit):

    • 0

      31
  • Exponent (8 bits):

    • 0

      30
    • 1

      29
    • 1

      28
    • 1

      27
    • 0

      26
    • 0

      25
    • 1

      24
    • 1

      23
  • Mantissa (23 bits):

    • 1

      22
    • 0

      21
    • 0

      20
    • 0

      19
    • 1

      18
    • 1

      17
    • 1

      16
    • 0

      15
    • 0

      14
    • 1

      13
    • 1

      12
    • 1

      11
    • 0

      10
    • 1

      9
    • 0

      8
    • 1

      7
    • 0

      6
    • 1

      5
    • 1

      4
    • 1

      3
    • 1

      2
    • 0

      1
    • 0

      0

Convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

A number in 32 bit single precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes 1 bit and it's either 0 for positive or 1 for negative numbers), exponent (8 bits), mantissa (23 bits)

Latest decimal numbers converted from base ten to 32 bit single precision IEEE 754 floating point binary standard representation

How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the base ten positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =


    1 - 1000 0011 - 100 1010 1100 0110 1010 0111